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13
Universal Gravitation
CHAPTER OUTLINE
13.1
Newton’s Law of Universal Gravitation
13.2
Free-Fall Acceleration and the Gravitational Force
13.3
Analysis Model: Particle in a Field (Gravitational)
13.4
Kepler’s Laws and the Motion of Planets
13.5
Gravitational Potential Energy
13.6
Energy Considerations in Planetary and Satellite Motion
* An asterisk indicates a question or problem new to this edition.
ANSWERS TO OBJECTIVE QUESTIONS
OQ13.1
Answer (c). Ten terms are needed in the potential energy:
U = U12 + U13 + U14 + U15 + U23 + U24 + U25 + U34 + U35 + U45
OQ13.2
The ranking is a > b = c. The gravitational potential energy of the
Earth-Sun system is negative and twice as large in magnitude as the
kinetic energy of the Earth relative to the Sun. Then the total energy
is negative and equal in absolute value to the kinetic energy.
OQ13.3
Answer (d). The satellite experiences a gravitational force, always
directed toward the center of its orbit, and supplying the centripetal
force required to hold it in its orbit. This force gives the satellite a
centripetal acceleration, even if it is moving with constant angular
speed. At each point on the circular orbit, the gravitational force is
directed along a radius line of the path, and is perpendicular to the
motion of the satellite, so this force does no work on the satellite.
OQ13.4
Answer (d). Having twice the mass would make the surface
gravitational field two times larger. But the inverse square law says
that having twice the radius would make the surface acceleration due
to gravitation four times smaller. Altogether, g at the surface of B
684
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Chapter 13
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becomes (2 m/s2)(2)/4 = 1 m/s2.
OQ13.5
Answer (b). Switching off gravity would let the atmosphere
evaporate away, but switching off the atmosphere has no effect on
the planet’s gravitational field.
OQ13.6
Answer (b). The mass of a spherical body of radius R and
density ρ is M = ρV = ρ(4πR3/3). The escape velocity from the
surface of this body may then be written in either of the following
equivalent forms:
vesc =
2GM
R
and vesc =
2G ⎛ 4πρR 3 ⎞
=
R ⎜⎝ 3 ⎟⎠
8πρGR 2
3
We see that the escape velocity depends on the three properties
(mass, density, and radius) of the planet. Also, the weight of an
object on the surface of the planet is Fg = mg = GMm/R2, giving
G ⎡ ⎛ 4π R 3 ⎞ ⎤ 4
g = GM R = 2 ⎢ ρ ⎜
⎥ = πρGR
R ⎣ ⎝ 3 ⎟⎠ ⎦ 3
2
The free-fall acceleration at the planet’s surface then depends on the
same properties as does the escape velocity. Changing the value of g
would necessarily change the escape velocity. Of the listed
quantities, the only one that does not affect the escape velocity is the
mass of the object.
OQ13.7
(i)
Answer (e). According to the inverse square law, 1/42 = 16 times
smaller.
(ii)
Answer (c). mv2/r = GMm/r2 predicts that v is proportional to
(1/r)1/2, so it becomes (1/4)1/2 = 1/2 as large.
(iii) Answer (a). According to Kepler’s third law, (43)1/2 = 8 times
larger; also, the circumference is 4 times larger and the speed
1/2 as large: 4/(1/2) = 8.
OQ13.8
Answer (b). The Earth is farthest from the sun around July 4 every
year, when it is summer in the northern hemisphere and winter in
the southern hemisphere. As described by Kepler’s second law, this
is when the planet is moving slowest in its orbit. Thus it takes more
time for the planet to plod around the 180° span containing the
minimum-speed point.
OQ13.9
The ranking is b > a > c = d > e. The force is proportional to the
product of the masses and inversely proportional to the square of the
separation distance, so we compute m1m2/r2 for each case: (a) 2·3/12 =
6 (b) 18 (c) 18/4 = 4.5 (d) 4.5 (e) 16/4 = 4.
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686
Universal Gravitation
OQ13.10
Answer (c). The International Space Station orbits just above the
atmosphere, only a few hundred kilometers above the ground. This
distance is small compared to the radius of the Earth, so the
gravitational force on the astronaut is only slightly less than on the
ground. We might think the gravitational force is zero or nearly zero,
because the orbiting astronauts appear to be weightless. They and the
space station are in free fall, so the normal force of the space station’s
wall/floor/ceiling on the astronauts is zero; they float freely around
the cabin.
OQ13.11
Answer (e). We assume that the elliptical orbit is so elongated that
the Sun, at one focus, is almost at one end of the major axis. If the
period, T, is expressed in years and the semimajor axis, a, in
astronomical units (AU), Kepler’s third law states that T 2 = a 3 .
Thus, for Halley’s comet, with a period of T = 76 y, the semimajor
axis of its orbit is
a=
3
(76)2
= 18 AU
The length of the major axis, and the approximate maximum distance
from the Sun, is 2a = 36 AU.
ANSWERS TO CONCEPTUAL QUESTIONS
CQ13.1
CQ13.2
CQ13.3
(a) The gravitational force is conservative. (b) Yes. An encounter with
a stationary mass cannot permanently speed up a spacecraft. But
Jupiter is moving. A spacecraft flying across its orbit just behind the
planet will gain kinetic energy because of the change in potential
energy of the spacecraft-planet system. This is a collision because the
spacecraft and planet exert forces on each other while they are
isolated from outside forces. It is an elastic collision because only
conservative forces are involved. (c) The planet loses kinetic energy
as the spacecraft gains it.
GM
Cavendish determined G. Then from g = 2 , one may determine
R
the mass of the Earth. The term “weighed” is better expressed as
“massed.”
For a satellite in orbit, one focus of an elliptical orbit, or the center of
a circular orbit, must be located at the center of the Earth. If the
satellite is over the northern hemisphere for half of its orbit, it must
be over the southern hemisphere for the other half. We could share
with Easter Island a satellite that would look straight down on
Arizona each morning and vertically down on Easter Island each
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Chapter 13
687
evening.
CQ13.4
(a)
Every point q on the sphere that does not lie along the axis
connecting the center of the sphere and the particle will have
companion point q' for which the components of the
gravitational force perpendicular to the axis will cancel. Point q'
can be found by rotating the sphere through 180° about the axis.
(b)
The forces will not necessarily cancel if the mass is not
uniformly distributed, unless the center of mass of the
nonuniform sphere still lies along the axis.
ANS. FIG. CQ13.4
CQ13.5
The angular momentum of a planet going around a sun is
conserved. (a) The speed of the planet is maximum at closest
approach. (b) The speed is a minimum at farthest distance. These two
points, perihelion and aphelion respectively, are 180° apart, at
opposite ends of the major axis of the orbit.
CQ13.6
Set the universal description of the gravitational force, Fg =
CQ13.7
(a)
In one sense, ‘no’. If the object is at the very center of the Earth
there is no other mass located there for comparison and the
formula does not apply in the same way it was being applied
while the object was some distance from the center. In another
sense, ‘yes’. One would have to compare, though, the distance
between the object with mass m to the other individual masses
that make up the Earth.
(b)
The gravitational force of the Earth on an object at its center
must be zero, not infinite as one interpretation of Equation 11.1
would suggest. All the bits of matter that make up the Earth
pull in different outward directions on the object, causing the
net force on it to be zero.
CQ13.8
GMX m
,
RX2
equal to the local description, Fg = magravitational, where Mx and Rx are
the mass and radius of planet X, respectively, and m is the mass of a
“test particle.” Divide both sides by m.
The escape speed from the Earth is 11.2 km/s and that from the
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Universal Gravitation
Moon is 2.3 km/s, smaller by a factor of 5. The energy required—and
fuel—would be proportional to v2, or 25 times more fuel is required
to leave the Earth versus leaving the Moon.
CQ13.9
Air resistance causes a decrease in the energy of the satellite-Earth
system. This reduces the radius of the orbit, bringing the satellite
closer to the surface of the Earth. A satellite in a smaller orbit,
however, must travel faster. Thus, the effect of air resistance is to
speed up the satellite!
SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 13.1
P13.1
Newton’s Law of Universal Gravitation
This is a direct application of the equation expressing Newton’s law of
gravitation:
(
(1.50 kg ) 15.0 × 10−3 kg
GMm
−11
2
2
F=
= 6.67 × 10 N ⋅ m /kg
2
r2
4.50 × 10−2 m
)
(
(
)
)
= 7.41 × 10−10 N
P13.2
For two 70-kg persons, modeled as spheres,
Fg =
Gm1m2 ( 6.67 × 10
=
r2
−11
N ⋅ m 2/kg 2 ) ( 70 kg ) ( 70 kg )
( 2 m )2
~ 10−7 N
P13.3
(a)
At the midpoint between the two objects, the forces exerted by the
200-kg and 500-kg objects are oppositely directed,
Gm1m2
r2
and from
Fg =
we have
∑F =
G ( 50.0 kg ) ( 500 kg − 200 kg )
( 2.00 m )
2
= 2.50 × 10−7 N
toward the 500-kg object.
(b) At a point between the two objects at a distance d from the
500-kg object, the net force on the 50.0-kg object will be zero
when
G ( 50.0 kg ) ( 200 kg )
( 4.00 m − d )
2
=
G ( 50.0 kg ) ( 500 kg )
d2
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Chapter 13
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To solve, cross-multiply to clear of fractions and take the square
root of both sides. The result is
d = 2.45 m from the 500-kg object toward the smaller object .
P13.4
(a)
The Sun-Earth distance is 1.496 × 1011 m and the Earth-Moon
distance is 3.84 × 108 m, so the distance from the Sun to the Moon
during a solar eclipse is
1.496 × 1011 m − 3.84 × 108 m = 1.492 × 1011 m
The mass of the Sun, Earth, and Moon are
MS = 1.99 × 1030 kg
ME = 5.98 × 1024 kg
and MM = 7.36 × 1022 kg
We have
FSM =
=
Gm1m2
r2
(6.67 × 10−11 N ⋅ m2 /kg 2 )(1.99 × 1030 kg )(7.36 × 1022 kg )
(1.492 × 10
11
m)
2
= 4.39 × 1020 N
(b)
FEM
(6.67 × 10
=
−11
N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg ) ( 7.36 × 1022 kg )
( 3.84 × 10
8
m)
2
= 1.99 × 1020 N
(c)
FSE =
(6.67 × 10
−11
N ⋅ m 2/kg 2 ) ( 1.99 × 1030 kg ) ( 5.98 × 1024 kg )
(1.496 × 10
11
m)
2
= 3.55 × 1022 N
(d)
The force exerted by the Sun on the Moon is much stronger than
the force of the Earth on the Moon. In a sense, the Moon orbits the
Sun more than it orbits the Earth. The Moon’s path is everywhere
concave toward the Sun. Only by subtracting out the solar orbital
motion of the Earth-Moon system do we see the Moon orbiting
the center of mass of this system.
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690
P13.5
Universal Gravitation
With one metric ton = 1 000 kg,
F = m1 g =
g=
Gm1m2
r2
Gm2 ( 6.67 × 10
=
r2
−11
N ⋅ m 2/kg 2 ) ( 4.00 × 107 kg )
(100 m )2
= 2.67 × 10−7 m/s 2
P13.6
The force exerted on the 4.00-kg mass by the
2.00-kg mass is directed upward and given
by

mm
F12 = G 22 1 ĵ
r12
= ( 6.67 × 10−11 N ⋅ m 2/kg 2 )
( 4.00 kg )( 2.00 kg ) ĵ
( 3.00 m )2
= 5.93 × 10
−11
ANS. FIG. P13.6
ĵ N
The force exerted on the 4.00-kg mass by the 6.00-kg mass is directed to
the left:

mm
F32 = G 2 2 3 − î
r32
( )
= ( −6.67 × 10−11 N ⋅ m 2/kg 2 )
( 4.00 kg )(6.00 kg ) î
( 4.00 m )2
= −10.0 × 10−11 î N
Therefore, the resultant force on the 4.00-kg mass is



F4 = F24 + F64 = −10.0î + 5.93 ĵ × 10−11 N (
*P13.7
)
The magnitude of the gravitational force is given by
Gm1m2 ( 6.672 × 10−11 N · m 2 /kg 2 ) ( 2.00 kg ) ( 2.00 kg )
F=
=
( 0.300 m )2
r2
= 2.97 × 10−9 N
P13.8
Gm1m2
, we
r2
5
would find that the mass of a sphere is 1.22 × 10 kg! If the spheres
have at most a radius of 0.500 m, the density of spheres would be at
Assume the masses of the sphere are the same. Using Fg =
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Chapter 13
691
least 2.34 × 105 kg/m3, which is ten times the density of the most dense
element, osmium.
The situation is impossible because no known element could compose
the spheres.
P13.9
We are given m1 + m2 = 5.00 kg, which means that m2 = 5.00 kg − m1.
Newton’s law of universal gravitation then becomes
m1m2
r2
⇒ 1.00 × 10−8 N
F=G
= ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
( 5.00 kg ) m1 − m
2
1
(1.00 × 10
=
−8
6.67 × 10
Thus,
m12 − ( 5.00 kg ) m1 + 6.00 kg = 0
or
( m1 − 3.00 kg )( m1 − 2.00 kg ) = 0
m1 ( 5.00 kg − m1 )
( 0.200 m )2
N ) ( 0.040 0 m 2 )
−11
N ⋅ m /kg
2
2
= 6.00 kg 2
giving m1 = 3.00 kg, so m2 = 2.00 kg . The answer m1 = 2.00 kg and
m2 = 3.00 kg is physically equivalent.
P13.10
Let θ represent the angle each cable makes with the
vertical, L the cable length, x the distance each ball is
displaced by the gravitational force, and d = 1 m the
original distance between them. Then r = d − 2x is the
separation of the balls. We have
∑ Fy = 0: T cosθ − mg = 0
∑ Fx = 0: T sin θ −
Then
tan θ =
L −x
ANS. FIG. P13.10
Gmm
r 2 mg
x
2
Gmm
=0
r2
2
=
Gm
2
g ( d − 2x )
→
Gm
2
x ( d − 2x ) = g L2 − x 2
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Universal Gravitation
Gm
is numerically small. We expect that x is very small
g
compared to both L and d, so we can treat the term (d − 2x) as d, and
(L2 − x2) as L2. We then have
The factor
x (1 m )
2
(6.67 × 10
=
−11
N ⋅ m 2/kg 2 ) ( 100 kg )
( 9.80 m/s )
2
( 45.00 m )
x = 3.06 × 10−8 m
Section 13.2
P13.11
Free-Fall Acceleration and the Gravitational Force
The distance of the meteor from the center of Earth is R + 3R = 4R.
Calculate the acceleration of gravity at this distance.
g=
GM (6.67 × 10−11 N ⋅ m 2 /kg 2 )(5.98 × 1024 kg)
=
r2
[4(6.37 × 106 m)]2
= 0.614 m/s 2 , toward Earth
P13.12
The gravitational field at the surface of the Earth or Moon is given by
GM
g= 2 .
R
The expression for density is
so
and
ρ =
M
M
=
,
V
4
π R3
3
4
M = π ρR 3
3
4
G ⎛ πρ R 3 ⎞
⎝3
⎠ 4
g=
= Gπρ R
2
R
3
Noting that this equation applies to both the Moon and the Earth, and
dividing the two equations,
4
Gπρ M RM
gM
ρ R
3
=
= M M
gE
4
ρE RE
GπρE RE
3
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Chapter 13
693
Substituting for the fractions,
()
1 ρM 1
=
6
ρE 4
P13.13
(a)
ρM 4
2
=
=
ρE
6
3
and
For the gravitational force on an object in the neighborhood of
Miranda, we have
mobj g =
GmobjmMiranda
2
rMiranda
(6.67 × 10
Gm
g = 2 Miranda =
rMiranda
−11
N ⋅ m 2 / kg 2 ) ( 6.68 × 1019 kg )
( 242 × 10
3
m)
2
= 0.076 1 m/s 2
(b)
We ignore the difference (of about 4%) in g between the lip and
the base of the cliff. For the vertical motion of the athlete, we have
1 2
ay t
2
1
−5 000 m = 0 + 0 + ( −0.076 1 m/s 2 ) t 2
2
y f = y i + vyi +
⎛ 2 ( 5 000 m ) s 2 ⎞
t=⎜
⎟
⎝ 0.076 1 m ⎠
(c)
1/2
= 363 s
1
x f = xi + vxit + axt 2 = 0 + ( 8.50 m/s )( 363 s ) + 0 = 3.08 × 103 m
2
We ignore the curvature of the surface (of about 0.7°) over the
athlete’s trajectory.
(d)
vxf = vxi = 8.50 m/s
vyf = vyi + ay t = 0 − ( 0.076 1 m/s 2 ) ( 363 s ) = −27.6 m/s
(
)

Thus v f = 8.50î − 27.6 ĵ m/s = 8.502 + 27.62 m/s at
⎛ 27.6 m/s ⎞
tan −1 ⎜
= 72.9° below the x axis.
⎝ 8.50 m/s ⎟⎠

v f = 28.9 m/s at 72.9° below the horizontal
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694
Universal Gravitation
Section 13.3
P13.14
(a)
Analysis Model: Particle in a Field (Gravitational)
g1 = g 2 =
MG
r + a2
2
g1y = − g 2 y
g y = g1y + g 2 y = 0
g1x = g 2 x = g 2 cos θ
cos θ =
(a
r
2
+ r2 )
1/2
ANS. FIG. P13.14
( )

g = 2g 2 x − î
or

g=
(r
2MGr
2
+ a2 )
32
toward the center of mass
(b)
At r = 0, the fields of the two objects are equal in magnitude and
opposite in direction, to add to zero.
(c)
As r → 0, 2MGr(r 2 + a 2 )−3/2 approaches 2MG(0)/a 3 = 0.
(d)
When r is much greater than a, the angles the field vectors make
with the x axis become smaller. At very great distances, the field
vectors are almost parallel to the axis; therefore, they begin to
look like the field vector from a single object of mass 2M.
(e)
As r becomes much larger than a, the expression approaches
2MGr(r 2 + 02 )−3/2 = 2MGr/r 3 = 2MG/r 2 as required.
P13.15
The vector gravitational field at point O is given by
(
Gm
Gm
 Gm
g = 2 î + 2 ĵ + 2 cos 45.0°î + sin 45.0 ĵ
l
l
2l
so
)
( )
1 ⎞
 Gm ⎛
g = 2 ⎜1+
⎟ î + ĵ or
l ⎝
2 2⎠
1⎞
 Gm ⎛
g = 2 ⎜ 2 + ⎟ toward the opposite corner.
l ⎝
2⎠
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Chapter 13
695
ANS. FIG. P13.15
P13.16
(a)
−11
2
2
30
3
GMm ( 6.67 × 10 N ⋅ m /kg ) ⎡⎣100 ( 1.99 × 10 kg ) ( 10 kg ) ⎤⎦
F=
=
2
r2
(1.00 × 104 m + 50.0 m )
= 1.31 × 1017 N
(b)
ΔF =
GMm GMm
− 2
2
rfront
rback
2
2
ΔF GM ( rback − rfront )
Δg =
=
2
2
m
rfront
rback
ANS. FIG. P13.16
Δg = ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
2
2
⎡⎣100 ( 1.99 × 1030 kg ) ⎤⎦ ⎡( 1.01 × 10 4 m ) − ( 1.00 × 10 4 m ) ⎤
⎣
⎦
2
2
4
4
(1.00 × 10 m ) (1.01 × 10 m )
Δg = 2.62 × 1012 N/kg
Section 13.4
P13.17
Kepler’s Laws and the Motion of Planets
The gravitational force on mass located at distance r from the center of
the Earth is Fg = mg = GME m/r 2 . Thus, the acceleration of gravity at
this location is g = GME /r 2 . If g = 9.00 m/s2 at the location of the
satellite, the radius of its orbit must be
GME
r=
=
g
(6.67 × 10
−11
N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )
9.00 m/s 2
= 6.66 × 106 m
From Kepler’s third law for Earth satellites, T 2 = 4π 2 r 3GMES, the
period is found to be
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696
Universal Gravitation
r3
T = 2π
= 2π
GME
(6.67 × 10
−11
(6.66 × 10
6
m)
3
N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )
= 5.41 × 103 s
or
⎛ 1h ⎞
T = ( 5.41 × 103 s ) ⎜
= 1.50 h = 90.0 min
⎝ 3 600 s ⎟⎠
P13.18
The gravitational force exerted by Jupiter on Io causes the centripetal
acceleration of Io. A force diagram of the satellite would show one
downward arrow.
∑ Fon Io = MIo a:
GMJ MIo MIo v 2 MIo ⎛ 2π r ⎞ 2 4π 2 rMIo
=
=
⎜
⎟ =
r2
r
r ⎝ T ⎠
T2
Thus the mass of Io divides out and we have Kepler’s third law with
m << M,
4π 2 r 3
4π 2 (4.22 × 108 m)3
⎛ 1d ⎞
MJ =
=
⎟
2
−11
2
2
2 ⎜
GT
(6.67 × 10 N ⋅ m /kg )(1.77 d) ⎝ 86 400 s ⎠
P13.19
2
and
M J = 1.90 × 1027 kg (approximately 316 Earth masses)
(a)
The desired path is an elliptical trajectory with the Sun at one of
the foci, the departure planet at the perihelion, and the target
planet at the aphelion. The perihelion distance rD is the radius of
the departure planet’s orbit, while the aphelion distance rT is the
radius of the target planet’s orbit. The semimajor axis of the
desired trajectory is then a = ( rD + rT )/2.
ANS. FIG. P13.19
If Earth is the departure planet, rD = 1.496 × 1011 m = 1.00 AU
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Chapter 13
697
With Mars as the target planet,
⎛
⎞
1 AU
rT = 2.28 × 1011 m ⎜
= 1.52 AU
11
⎝ 1.496 × 10 m ⎟⎠
Thus, the semimajor axis of the minimum energy trajectory is
a=
rD + rT 1.00 AU + 1.52 AU
=
= 1.26 AU
2
2
Kepler’s third law, T2 = a3, then gives the time for a full trip
around this path as
T = a3 =
(1.26 AU )3
= 1.41 yr
so the time for a one-way trip from Earth to Mars is
Δt =
P13.20
1.41 yr
1
T=
= 0.71 yr
2
2
(b)
This trip cannot be taken at just any time. The departure must
be timed so that the spacecraft arrives at the aphelion when the
target planet is located there.
(a)
The particle does possess angular momentum, because it is
not headed straight for the origin.
(b)
Its angular momentum is constant. There are no identified
outside influences acting on the object.
(c)
Since speed is constant, the distance traveled between tA and
tB is equal to the distance traveled between tC and tD . The area
of a triangle is equal to one-half its (base) width across one side
times its (height) dimension perpendicular to that side.
So
1
1
bv0 ( tB − tA ) = bv0 ( tD − tC )
2
2
states that the particle’s radius vector sweeps out equal areas in
equal times.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
698
P13.21
Universal Gravitation
Applying Newton’s second law, ∑ F = ma yields
Fg = mac for each star:
GMM Mv 2
=
r
( 2r )2
or
M=
4v 2 r
G
We can write r in terms of the period, T, by
considering the time and distance of one complete
cycle. The distance traveled in one orbit is the
circumference of the stars’ common orbit, so
2π r = vT. Therefore,
M=
ANS. FIG. P13.21
4v 2 r 4v 2 ⎛ vT ⎞
=
⎜
⎟
G
G ⎝ 2π ⎠
3
2v 3T 2 ( 220 × 10 m/s ) ( 14.4 d ) ( 86 400 s/d )
M=
=
πG
π ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )
3
so,
= 1.26 × 1032 kg = 63.3 solar masses
P13.22
To find the angular displacement of planet Y, we
apply Newton’s second law:
∑ F = ma:
Gmplanet Mstar
r2
=
mplanet v 2
r
Then, using v = rω ,
GMstar
= v 2 = r 2ω 2
r
GMstar = r 3ω 2 = rx3ω x2 = ry3ω y2
solving for the angular velocity of planet Y gives
⎛r ⎞
ωy = ωx ⎜ x ⎟
⎝ ry ⎠
32
⎛ 90.0° ⎞ 3 2
468°
=⎜
3 =
⎟
5.00 yr
⎝ 5.00 yr ⎠
ANS. FIG. P13.22
So, given that there are 360° in one revolution we
convert 468° to find that planet Y has turned through
1.30 revolutions .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.23
699
By Kepler’s third law, T2 = ka3 (a = semimajor
axis). For any object orbiting the Sun, with T in
years and a in AU, k = 1.00. Therefore, for
Comet Halley, and suppressing units,
⎛ 0.570 + y ⎞
(75.6) = (1.00) ⎜
⎟⎠
⎝
2
3
2
The farthest distance the comet gets from the
Sun is
y = 2 ( 75.6 )
23
ANS. FIG. P13.23
− 0.570 = 35.2 AU
(out around the orbit of Pluto).
*P13.24
By conservation of angular momentum for the satellite, rp v p = ra va , or
vp
va
=
ra 2 289 km + 6.37 × 103 km 8 659 km
=
=
= 1.27
rp
459 km + 6.37 × 103 km
6 829 km
We do not need to know the period.
P13.25
For an object in orbit about Earth, Kepler’s third law gives the relation
between the orbital period T and the average radius of the orbit
(“semimajor axis”) as
⎛ 4π 2 ⎞ 3
T2 = ⎜
r
⎝ GME ⎟⎠
We assume that the two given distances in the problem statements are
the perigee and apogee, respectively.
Thus, if the average radius is
rmin + rmax 6 670 km + 385 000 km
=
2
2
5
= 1.96 × 10 km = 1.96 × 108 m
r=
The period (time for a round trip from Earth to the Moon) would be
r3
T = 2π
GME
= 2π
(6.67 × 10
−11
(1.96 × 10
8
m)
3
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg )
= 8.63 × 105 s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
700
Universal Gravitation
The time for a one-way trip from Earth to the Moon is then
⎛ 8.63 × 105 s ⎞ ⎛ 1 day ⎞
1
Δt = T = ⎜
⎟⎠ ⎜⎝ 8.64 × 10 4 s ⎟⎠ = 4.99 d
⎝
2
2
P13.26
The gravitational force on a small parcel of material at the star’s
equator supplies the necessary centripetal acceleration:
GMs m mv 2
=
= mRsω 2
Rs2
Rs
so
(6.67 × 10
GMs
ω=
=
Rs3
−11
N ⋅ m 2 /kg 2 ) ⎡⎣ 2 ( 1.99 × 1030 kg ) ⎤⎦
(10.0 × 10 m )
3
3
ω = 1.63 × 10 4 rad/s
P13.27
We find the satellite’s altitude from
GM J
(R
J
+d
)
=
2
(
4π 2 RJ + d
T
2
)
where d is the altitude of the satellite above Jupiter’s cloud tops. Then,
(
GM JT 2 = 4π 2 RJ + d
(6.67 × 10
−11
)
3
N ⋅ m 2 /kg 2 ) ( 1.90 × 1027 kg ) ( 9.84 × 3 600 )
2
= 4π 2 ( 6.99 × 107 + d )
3
which gives
d = 8.92 × 107 m = 89 200 km above the planet
P13.28
(a)
In T2 = 4 π2a3/GMcentral we take a = 3.84 × 108 m.
Mcentral =
=
4π 2 a 3
GT 2
4π 2 (3.84 × 108 m)3
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(27.3 × 86 400 s)2
= 6.02 × 1024 kg
This is a little larger than 5.98 × 1024 kg.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
(b)
701
The Earth wobbles a bit as the Moon orbits it, so both objects
move nearly in circles about their center of mass, staying on
opposite sides of it. The radius of the Moon’s orbit is therefore
a bit less than the Earth-Moon distance.
P13.29
The speed of a planet in a circular orbit is given by
∑ F = ma:
(a)
GMsun m mv 2
=
r2
r
→ v=
GMsun
r
For Mercury, the speed is
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
vM =
5.79 × 1010 m
= 4.79 × 10 4 m/s
and for Pluto,
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1.99 × 1030 kg)
5.91 × 1012 m
= 4.74 × 103 m/s
vp =
With greater speed, Mercury will eventually move farther from
the Sun than Pluto.
(b)
With original distances rP and rM perpendicular to their lines of
motion, they will be equally far from the Sun at time t, where
2 2
rP2 + vP2 t 2 = rM2 + v M
t
2
rP2 − rM2 = ( v M
− vP2 ) t 2
t=
=
( 5.91 × 10
( 4.79 × 10
4
12
m ) − ( 5.79 × 1010 m )
2
2
m/s ) − ( 4.74 × 103 m/s )
2
2
3.49 × 1025 m 2
= 1.24 × 108 s = 3.93 yr
2.27 × 109 m 2/s 2
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
702
Universal Gravitation
Section 13.5
P13.30
(a)
Gravitational Potential Energy
We compute the gravitational potential energy of the satelliteEarth system from
U=−
=−
GME m
r
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg )(100 kg )
( 6.37 + 2.00 ) × 106 m
= −4.77 × 109 J
(b), (c) The satellite and Earth exert forces of equal magnitude on each
other, directed downward on the satellite and upward on Earth.
The magnitude of this force is
F=
=
GME m
r2
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg )(100 kg )
( 8.37 × 10
6
m)
2
= 569 N
P13.31
The work done by the Moon’s gravitational field is equal to the
negative of the change of potential energy of the meteor-Moon system:
⎛ −Gm1m2
⎞
Wint = −ΔU = − ⎜
− 0⎟
⎝
⎠
r
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.36 × 1022 kg)(1.00 × 103 kg)
Wint =
1.74 × 106 m
= 2.82 × 109 J
*P13.32
The enery required is equal to the change in gravitational potential
energy of the object-Earth system:
U = −G
GME
Mm
and g =
so that
RE2
r
1⎞ 2
⎛ 1
ΔU = −GMm ⎜
− ⎟ = mgRE
⎝ 3RE RE ⎠ 3
2
ΔU = ( 1 000 kg ) ( 9.80 m s 2 ) ( 6.37 × 106 m ) = 4.17 × 1010 J
3
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Chapter 13
P13.33
(a)
703
The definition of density gives
3 ( 1.99 × 1030 kg )
MS
ρ=
=
= 1.84 × 109 kg/m 3
4 2 4π ( 6.37 × 106 m )3
πr
3 E
(b)
For an object of mass m on its surface, mg = GMS m/RE2. Thus,
−11
2
2
30
GMS ( 6.67 × 10 N ⋅ m /kg ) ( 1.99 × 10 kg )
g= 2 =
2
rE
(6.37 × 106 m )
= 3.27 × 106 m/s 2
(c)
Relative to Ug = 0 at infinity, the potential energy of the object-star
system at the surface of the white dwarf is
GMSm
rE
Ug = −
(6.67 × 10
=−
−11
N ⋅ m 2 /kg 2 ) ( 1.99 × 1030 kg ) ( 1.00 kg )
6.37 × 106 m
= − 2.08 × 1013 J
P13.34
(a)
Energy conservation of the object-Earth system from release to
radius r:
(K + U )
g
altitude h
(
= K + Ug
)
radius r
GME m 1
GME m
0−
= mv 2 −
RE + h 2
r
⎡
⎛1
1 ⎞⎤
v = ⎢ 2GME ⎜ −
⎥
⎝ r RE + h ⎟⎠ ⎦
⎣
(b)
f
f
i
i
∫ dt = ∫ −
1/2
=−
dr
dt
i
dr
dr
= ∫ . The time of fall is, suppressing units,
v
v
f
Δt =
RE +h
∫
RE
⎡
⎛1
1 ⎞⎤
⎢ 2GME ⎜ −
⎥
⎝ r RE + h ⎟⎠ ⎦
⎣
−1/2
dr
Δt = ( 2 × 6.67 × 10−11 × 5.98 × 1024 )
6.87×106 m
−1/2
1
⎡⎛ 1
⎞⎤
×
∫ 6 ⎢⎣⎜⎝ r − 6.87 × 106 m ⎟⎠ ⎥⎦
6.37×10 m
−1/2
dr
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
704
Universal Gravitation
We can enter this expression directly into a mathematical
calculation program.
Alternatively, to save typing we can change variables to u =
r
.
106
Then
Δt = ( 7.977 × 10
)
14 −1/2
6.87
1
⎛ 1
⎞
∫6.37 ⎜⎝ 106 u − 6.87 × 106 ⎟⎠
106
6.87
1 ⎞
⎛1
= 3.541 × 10
⎜ −
⎟
6 −1/2 ∫ ⎝ u
(10 ) 6.37 6.87 ⎠
−8
−1/2
106 du
−1/2
du
A mathematics program returns the value 9.596 for this integral,
giving for the time of fall
Δt = 3.541 × 10−8 × 109 × 9.596 = 339.8 = 340 s
P13.35
(a)
Since the particles are located at the corners of an equilateral
triangle, the distances between all particle pairs is equal to
0.300 m. The gravitational potential energy of the system is then
⎛ Gm1m2 ⎞
U Tot = U 12 + U 13 + U 23 = 3U 12 = 3 ⎜ −
r12 ⎟⎠
⎝
U Tot = −
3 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.00 × 10−3 kg )
2
0.300 m
= −1.67 × 10−14 J
(b)
Section 13.6
P13.36
Each particle feels a net force of attraction toward the midpoint
between the other two. Each moves toward the center of the
triangle with the same acceleration. They collide simultaneously
at the center of the triangle.
Energy Considerations in
Planetary and Satellite Motion
We use the isolated system model for energy:
Ki + Ui = Kf + Uf
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Chapter 13
705
⎛ 1 1⎞ 1
1
mvi2 + GME m ⎜ − ⎟ = mv 2f
2
⎝ rf ri ⎠ 2
which becomes
⎛
1 2
1⎞ 1
vi + GME ⎜ 0 − ⎟ = v 2f
2
RE ⎠ 2
⎝
2GME
RE
or
v 2f = v12 −
and
⎛
2GME ⎞
v f = ⎜ v12 −
RE ⎟⎠
⎝
1/2
⎧⎪
⎡ 2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ⎤ ⎫⎪
2
4
v f = ⎨( 2.00 × 10 m/s ) − ⎢
⎥⎬
6.37 × 106 m
⎢⎣
⎥⎦ ⎪⎭
⎩⎪
1/2
= 1.66 × 10 4 m/s
*P13.37
To determine the energy transformed to internal energy, we begin by
calculating the change in kinetic energy of the satellite. To find the
initial kinetic energy, we use
vi2
GME
=
RE + h ( RE + h )2
which gives
Ki =
=
1
1 ⎛ GME m ⎞
mvi2 = ⎜
2
2 ⎝ RE + h ⎟⎠
−11
2
2
24
1 ⎡ ( 6.67 × 10 N ⋅ m kg ) ( 5.98 × 10 kg ) ( 500 kg ) ⎤
⎥
2 ⎢⎣
6.37 × 106 m + 0.500 × 106 m
⎦
= 1.45 × 1010 J
Also,
Kf =
1
1
2
mv 2f = ( 500 kg ) ( 2.00 × 103 m s ) = 1.00 × 109 J.
2
2
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
706
Universal Gravitation
The change in gravitational potential energy of the satellite-Earth
system is
⎛ 1
GME m GME m
1 ⎞
−
= GME m ⎜ −
⎟
Ri
Rf
⎝ Ri R f ⎠
ΔU =
= ( 6.67 × 10−11 N ⋅ m 2 kg 2 ) ( 5.98 × 1024 kg ) ( 500 kg )
× ( −1.14 × 10−8 m −1 )
= −2.27 × 109 J
The energy transformed into internal energy due to friction is then
ΔEint = K i − K f − ΔU = ( 14.5 − 1.00 + 2.27 ) × 109 J
= 1.58 × 1010 J
P13.38
To obtain the orbital velocity, we use
or
∑F =
mMG mv 2
=
R2
R
v=
MG
R
We can obtain the escape velocity from
1
mMG
2
mvesc
=
2
R
or
P13.39
(a)
2MG
=
R
vesc =
2v
The total energy of the satellite-Earth system at a given orbital
altitude is given by
Etot = −
GMm
2r
The energy needed to increase the satellite’s orbit is then,
suppressing units,
ΔE =
GMm ⎛ 1 1 ⎞
−
2 ⎜⎝ ri rf ⎟⎠
(6.67 × 10 )( 5.98 × 10 ) 10
=
−11
24
2
⎞
kg ⎛
1
1
−
⎜
10 m ⎝ 6 370 + 100 6 370 + 200 ⎟⎠
3
3
ΔE = 4.69 × 108 J = 469 MJ
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
(b)
707
Both in the original orbit and in the final orbit, the total energy
is negative, with an absolute value equal to the positive kinetic
energy. The potential energy is negative and twice as large as
the total energy. As the satellite is lifted from the lower to the
higher orbit, the gravitational energy increases, the kinetic energy
decreases, and the total energy increases. The value of each
becomes closer to zero. Numerically, the gravitational energy
increases by 938 MJ, the kinetic energy decreases by 469 MJ,
and the total energy increases by 469 MJ.
P13.40
(a)
The major axis of the orbit is 2a = 50.5 AU so a = 25.25 AU.
Further, in the textbook’s diagram of an ellipse, a + c = 50 AU,
so c = 24.75 AU. Then
e=
(b)
c 24.75
=
= 0.980
a 25.25
In T2 = Ks a3 for objects in solar orbit, the Earth gives us
(1 yr )
2
= K s ( 1 AU ) K s =
3
(1 yr )2
(1 AU )3
Then
T =
2
(c)
U=−
=−
(1 yr )2 ( 25.25 AU )3
3
(1 AU )
→ T = 127 yr
GMm
r
(6.67 × 10−11 N ⋅ m2 / kg 2 )(1.99 × 1030 kg )(1.20 × 1010 kg )
50 ( 1.496 × 1011 m )
= −2.13 × 1017 J
*P13.41
For her jump on Earth,
1
mvi2 = mgy f
2
[1]
which gives
vi = 2gy f = 2 ( 9.80 m/s ) ( 0.500 m ) = 3.13 m/s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
708
Universal Gravitation
We assume that she has the same takeoff speed on the asteroid. Here
1
GMA m
mvi2 −
= 0+0
2
RA
[2]
The equality of densities between planet and asteroid,
ρ=
ME
MA
=
4
4
π RE3
π RA3
3
3
implies
3
⎛R ⎞
MA = ⎜ A ⎟ ME
⎝ RE ⎠
[3]
Note also at Earth’s surface
g=
GME
RE2
[4]
Combining the equations [2], [1], [3], and [4] by substitution gives
1 2 GMA
v =
2 i
RA
GME
GME RA2
y
=
RE2 f
RE3
RA2 = y f RE = ( 0.500 m ) ( 6.37 × 106 m )
RA = 1.78 × 103 m
*P13.42
For a satellite in an orbit of radius r around the Earth, the total energy
GME
of the satellite-Earth system is E = −
. Thus, in changing from a
2r
circular orbit of radius r = 2RE to one of radius r = 3RE, the required
work is
W = ΔE = −
GME m GME m
1 ⎤
GME m
⎡ 1
+
= GME m ⎢
−
=
⎥
2rf
2ri
12RE
⎣ 4RE 6RE ⎦
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
*P13.43
(a)
709
The work must provide the increase in gravitational energy:
W = ΔU g = U gf − U gi
=−
=−
GME Mp
rf
GME Mp
RE + y
+
+
GME Mp
ri
GME Mp
RE
⎛ 1
1 ⎞
= GME Mp ⎜
−
⎝ RE RE + y ⎟⎠
⎛ 6.67 × 10−11 N ⋅ m 2 ⎞
=⎜
5.98 × 1024 kg ) ( 100 kg )
(
2
⎟
⎝
⎠
kg
×
(
1
1
−
6
6.37 × 10 m 7.37 × 106 m
)
W = 850 MJ
(b)
In a circular orbit, gravity supplies the centripetal force:
GME M p
( RE + y )2
=
Mp v 2
RE + y
Then,
1
1 GME Mp
Mp v 2 =
2
2 ( RE + y )
So, additional work = kinetic energy required is
1 ( 6.67 × 10
ΔW =
2
−11
N ⋅ m 2 / kg 2 ) ( 5.98 × 1024 kg ) ( 100 kg )
7.37 × 106 m
= 2.71 × 109 J
P13.44
(a)
The escape velocity from the solar system, starting at Earth’s
orbit, is given by
vsolar escape =
=
2MSunG
RSun
2 ( 1.99 × 1030 kg ) ( 6.67 × 10−11 N ⋅ m 2 / kg 2 )
1.50 × 109 m
= 42.1 km/s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
710
Universal Gravitation
(b)
Let x represent the variable distance from the Sun. Then,
2MSunG
x
v=
→ x=
v2
2MSunG
125 000 km
= 34.7 m/s, then
3 600 s
If v =
v2
( 34.7 m/s )
x=
=
30
2MSunG 2 ( 1.99 × 10 kg ) ( 6.67 × 10−11 N ⋅ m 2 / kg 2 )
2
= 2.20 × 1011 m
Note that at or beyond the orbit of Mars, 125 000 km/h is
sufficient for escape.
P13.45
Fc = FG gives
mv 2 GmME
=
r
r2
GME
r
which reduces to v =
and period =
(a)
2π r
r
= 2π r
.
v
GME
r = RE + 200 km = 6 370 km + 200 km = 6 570 km
Thus,
period = 2π ( 6.57 × 106 m )
×
(6.67 × 10
−11
6.57 × 106 m
N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )
T = 5.30 × 103 s = 88.3 min = 1.47 h
(b)
v=
=
GME
r
(6.67 × 10
−11
N ⋅ m 2/kg 2 ) ( 5.98 × 1024 kg )
6.57 × 106 m
= 7.79 km/s
(c)
K f + U f = K i + U i + energy input gives
input =
1 2 1 2 ⎛ −GME m ⎞ ⎛ −GME m ⎞
mv f − mvi + ⎜
⎟ −⎜
⎟⎠
2
2
rf
ri
⎝
⎠ ⎝
[1]
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
711
ri = RE = 6.37 × 106 m
2π RE
vi =
= 4.63 × 102 m/s
86 400 s
Substituting the appropriate values into [1] yields:
minimum energy input = 6.43 × 109 J
P13.46
The gravitational force supplies the needed centripetal acceleration.
GME m
Thus,
(a)
( RE + h)
2
=
mv 2
RE + h
2π r 2π ( RE + h )
T=
=
= 2π
GME
v
or
v2 =
GME
RE + h
( RE + h)3
( RE + h)
GME
GME
RE + h
(b)
v=
(c)
Minimum energy input is
(
) (
ΔEmin = K f + U gf − K i − U gi
)
This choice has the object starting with energy
Ki =
with vi =
1
mvi2
2
2π RE
2π RE
=
1.00 day 86 400 s
and
U gi = −
GME m
.
RE
Thus,
ΔEmin =
or
P13.47
(a)
1 ⎛ GME ⎞ GME m 1 ⎡ 4π 2 RE2 ⎤ GME m
m
−
− m⎢
⎥+
2 ⎜⎝ RE + h ⎟⎠ RE + h 2 ⎢⎣ ( 86 400 s )2 ⎥⎦
RE
⎡ RE + 2h ⎤
2π 2 RE2 m
ΔEmin = GME m ⎢
−
⎥
2 .
⎣ 2RE ( RE + h ) ⎦ ( 86 400 s )
Gravitational screening does not exist. The presence of the
satellite has no effect on the force the planet exerts on the
rocket.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
712
Universal Gravitation
(b) The rocket has a gravitational potential energy with respect to
Ganymede
6.67 × 10−11 N ⋅ m 2 ) m2 ( 1.495 × 1023 kg )
(
Gm1m2
U1 = −
=−
r
( 2.64 × 106 m ) kg 2
U 1 = ( −3.78 × 106 m 2 /s 2 ) m2
The rocket’s gravitational potential energy with respect to
Jupiter at the distance of Ganymede is
6.67 × 10−11 N ⋅ m 2 ) m2 ( 1.90 × 1027 kg )
(
Gm1m2
U2 = −
=−
r
(1.071 × 109 m ) kg 2
U 2 = ( −1.18 × 108 m 2 / s 2 ) m2
To escape from both requires
1
2
m2 vesc
= + ⎡⎣( 3.78 × 106 + 1.18 × 108 ) m 2 /s 2 ⎤⎦ m2
2
vesc = 2 × ( 1.22 × 108 m 2 / s 2 ) = 15.6 km/s
P13.48
(a)
For the satellite ∑ F = ma;
⎛ GME ⎞
vi = ⎜
⎝ r ⎟⎠
(b)
GmME mvi2
gives
=
r2
r
1/2
Conservation of momentum in the forward direction for the
exploding satellite gives:
( ∑ mv )i = ( ∑ mv ) f
5mvi = 4mv + m0
5
5 ⎛ GME ⎞
v = vi =
⎜
⎟
4
4⎝ r ⎠
(c)
1/2
With velocity perpendicular to radius, the orbiting fragment is at
perigee. Its apogee distance and speed are related to r and v
by 4mrv = 4mrf vf and
1
GME 4m 1
GME 4m
4mv 2 −
= 4mv 2f −
2
r
2
rf
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
Substituting v f =
713
vr
we have
rf
1 2 GME 1 v 2 r 2 GME
v −
=
−
2
r
2 rf2
rf
Further, substituting v 2 =
25 GME
gives
16 r
25 GME GME 25 GME r GME
−
=
−
32 r
r
32 rf2
rf
−7
25r
1
=
−
2
32r 32rf rf
Clearing fractions we have −7rf2 = 25r 2 − 32rrf , or
2
⎛ rf ⎞
⎛ rf ⎞
7 ⎜ ⎟ − 32 ⎜ ⎟ + 25 = 0
⎝r⎠
⎝r⎠
giving
rf
r
=
+32 ± 32 2 − 4 ( 7 ) ( 25 )
14
=
50
14
or
.
14
14
The latter root describes the starting point.
The outer end of the orbit has
*P13.49
rf
r
=
25
:
7
rf =
25r
7
The height attained is not small compared to the radius of the Earth, so
GM1 M2
U = mgy does not apply; U = −
does. From launch to apogee at
r
height h, conservation of energy gives
K i + U i + ΔEmech = K f + U f
GME Mp
GME Mp
1
Mp vi2 −
+0= 0−
2
RE
RE + h
The mass of the projectile cancels out, giving
1 2 GME GME
v −
=
2 i
RE
RE + h
GME
RE + h =
1 2 GME
v −
2 i
RE
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
714
Universal Gravitation
h=
=
GME
− RE
1 2 GME
vi −
2
RE
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg )
6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg )
(
1
2
3
(10.0 × 10 m/s ) −
2
(6.37 × 106 m )
− 6.37 × 106 m
= 2.52 × 107 m
Additional Problems
P13.50
(a)
When the rocket engine shuts off at an altitude of 250 km, we may
consider the rocket to be beyond Earth’s atmosphere. Then, its
mechanical energy will remain constant from that instant until it
comes to rest momentarily at the maximum altitude. That is,
KE f + PE f = KEi + PEi , or
0−
GME m 1
GME m
= m vi2 −
rmax
2
ri
or
1
rmax
=−
vi2
1
+
2GME ri
With rl = RE + 250 km = 6.37 × 106 m + 250 × 103 m = 6.62 × 106 m
and vi = 6.00 km/s = 6.00 × 103 m/s, this gives
1
rmax
=−
2 ( 6.67 × 10
(6.00 × 10
−11
3
m/s )
N ⋅ m /kg
2
2
2
)( 5.98 × 10
24
kg )
+
1
6.62 × 106 m
= 1.06 × 10−7 m −1
or rmax = 9.44 × 106 m. The maximum distance from Earth’s surface
is then
hmax = rmax – RE = 9.44 × 106 m – 6.37 × 106 m = 3.07 × 106 m
(b)
If the rocket were fired from a launch site on the equator, it would
have a significant eastward component of velocity because of the
Earth’s rotation about its axis. Hence, compared to being fired
from the South Pole, the rocket’s initial speed would be greater,
and the rocket would travel farther from Earth .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.51
715
For a 6.00-km diameter cylinder, r = 3 000 m and to simulate
1g = 9.80 m/s2,
v2
= ω 2r
r
g
ω=
= 0.057 2 rad/s
r
g=
The required rotation rate of the cylinder is
1 rev
.
110 s
(For a description of proposed cities in space, see Gerard K. O’Neill
in Physics Today, Sept. 1974. and the Wikipedia article on “Rotating
Wheel Space Station at
http://en.wikipedia.org/wiki/Rotating_wheel_space_station)
*P13.52
To approximate the height of the sulfur, set
h = 70 000 m and g Io =
mv 2
= mg Io h, with
2
GM
= 1.79 m s 2 . This gives
2
r
v = 2g Io h = 2 ( 1.79 m/s 2 ) ( 70 000 m )
≈ 500 m s ( over 1 000 mi h )
We can obtain a more precise answer from conservation of energy:
1 2 GMm
GMm
mv −
=−
2
r1
r2
1 2
v = ( 6.67 × 10−11 N ⋅ m 2 kg 2 ) ( 8.90 × 1022 kg )
2
1
1
×
−
6
1.82 × 10 m 1.89 × 106 m
(
)
v = 492 m/s
*P13.53
(a)
The radius of the satellite’s orbit is
r = RE + h = 6.37 × 106 m + 2.80 × 106 m = 9.17 × 106 m
Then, modifying Kepler’s third law for orbital motion about the
Earth rather than the Sun, we have
4π 2 ( 9.17 × 106 m )
⎛ 4π 2 ⎞ 3
T =⎜
r =
⎝ GME ⎟⎠
(6.67 × 10−11 N ⋅ m2 kg 2 )( 5.98 × 1024 kg )
3
2
= 7.63 × 107 s 2
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
716
Universal Gravitation
or
(b)
⎛ 1h ⎞
T = ( 8.74 × 103 s ) ⎜
= 2.43 h
⎝ 3 600 s ⎟⎠
The constant tangential speed of the satellite is
6
2π r 2π ( 9.17 × 10 m )
v=
=
= 6.60 × 103 m/s = 6.60 km/s
3
T
8.74 × 10 s
(c)
The satellite’s only acceleration is centripetal acceleration, so
3
v 2 ( 6.60 × 10 m/s )
a = ac =
=
= 4.74 m/s 2 toward the Earth
6
r
9.17 × 10 m
2
GM
and the relation v = (2π r/T), one finds
r
the radius of the orbit to be smaller than the radius of the Earth, so the
spacecraft would need to be in orbit underground.
P13.54
If one uses the result v =
P13.55
The acceleration of an object at the center of
the Earth due to the gravitational force of the
M
Moon is given by a = G 2M .
d
At the point A nearest the Moon,
a+ = G
MM
ANS. FIG. P13.55
( d − RE )2
At the point B farthest from the Moon,
a− = G
MM
( d + RE )2
From the above, we have
⎤
Δg M ( a+ − a− ) GM M ⎡
1
1
=
=
−
⎢
⎥
g
g
g ⎢⎣ ( d − RE )2 ( d + RE )2 ⎥⎦
Evaluating this expression, we find across the planet
−11
2
2
22
Δg M ( 6.67 × 10 N ⋅ m kg ) ( 7.36 × 10 kg )
=
g
9.80 m/s 2
⎡
⎤
1
1
⎢
⎥
×
2 −
2
⎢ ( 3.84 × 108 m − 6.37 × 106 m ) ( 3.84 × 108 m + 6.37 × 106 m ) ⎥
⎣
⎦
= 2.25 × 10−7
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.56
(a)
717
The only force acting on the astronaut is the normal force exerted
on him by the “floor” of the cabin. The normal force supplies the
centripetal force:
Fc =
mv 2
r
and
n=
mg
2
This gives
mv 2 mg
=
→v=
r
2
v=
gr
2
(9.80 m/s 2 )(10.0 m)
→ v = 7.00 m/s
2
Since v = rω , we have
ω=
v 7.00 m/s
=
= 0.700 rad/s
r
10.0 m
ANS. FIG. P13.56
(b)
Because his feet stay in place on the floor, his head will be
moving at the same tangential speed as his feet. However,
his feet and his head are travelling in circles of different radii.
(c)
If he stands up without holding on to anything with his hands,
the only force on his body is radial. Because the wall of the
cabin near the traveler's head moves in a smaller circle, it
moves at a slower tangential speed than that of the traveler's
head so his head moves toward the wall—if he is not careful,
there could be a collision. This is an example of the Coriolis
force investigated in Section 6.3. Holding onto to a rigid support
with his hands will provide a tangential force to the traveler to
slow the upper part of his body down.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
718
P13.57
Universal Gravitation
(a)
Ignoring air resistance, the energy conservation for the objectEarth system from firing to apex is given by,
(K + U ) = (K + U )
g
g
i
f
1
GmME
GmME
mvi2 −
= 0−
2
RE
RE + h
where
1 2
GmME
. Then
mvesc =
2
RE
1 2 1 2
1 2 RE
vi − vesc = − vesc
2
2
2
RE + h
2
vesc
− vi2 =
2
vesc
RE
RE + h
1
R +h
= 2E
2
v − vi
vesc RE
2
esc
h=
2
2
2
vesc
RE
vesc
RE − vesc
RE + vi2 RE
−
R
=
E
2
2
vesc
− vi2
vesc
− vi2
h=
RE vi2
2
vesc
− vi2
h=
(b)
(6.37 × 10
6
m ) ( 8.76 km/s )
2
(11.2 km/s ) − ( 8.76 km/s )
2
2
= 1.00 × 107 m
The fall of the meteorite is the time-reversal of the upward flight
of the projectile, so it is described by the same energy equation:
RE ⎞
2 ⎛
vi2 = vesc
1
−
⎜⎝
RE + h ⎟⎠
h ⎞
2 ⎛
= vesc
⎜⎝ R + h ⎟⎠
E
⎞
2⎛
2.51 × 107 m
= ( 11.2 × 103 m/s ) ⎜
6
7
⎝ 6.37 × 10 m + 2.51 × 10 m ⎟⎠
= 1.00 × 108 m 2 / s 2
vi = 1.00 × 10 4 m/s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.58
(a)
719
Ignoring air resistance, the energy conservation for the objectEarth system from firing to apex is given by,
(K + U ) = (K + U )
g
g
i
f
1
GmME
GmME
mvi2 −
= 0−
2
RE
RE + h
where
1 2
GmME
mvesc =
. Then
2
RE
1 2 1 2
1 2 RE
vi − vesc = − vesc
2
2
2
RE + h
2
vesc
− vi2 =
2
vesc
RE
RE + h
1
R +h
= 2E
2
v − vi
vesc RE
2
esc
h=
h=
(b)
2
2
2
vesc
RE
vesc
RE − vesc
RE + vi2 RE
−
R
=
E
2
2
vesc
− vi2
vesc
− vi2
RE vi2
2
vesc
− vi2
The fall of the meteorite is the time-reversal of the upward flight
of the projectile, so it is described by the same energy equation.
From (a) above, replacing vi with vf , we have
2
2
v 2f = vesc
− vesc
RE
RE + h
RE ⎞
2 ⎛
v 2f = vesc
⎜⎝ 1 − R + h ⎟⎠
E
v f = vesc
(c)
h
RE + h
With vi << vesc , h ≈
GME
RE vi2 RE vi2 RE
vi2
g
=
.
But
,
so
, in
=
h=
2
RE2
vesc
2GME
2g
agreement with 02 = vi2 + 2 ( − g ) ( h − 0 ) .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
720
P13.59
Universal Gravitation
(a)
Let R represent the radius of the asteroid. Then its volume is
4
4
π R 3 and its mass is ρ π R 3 . For your orbital motion, ∑ F = ma
3
3
gives
Gm1m2 m2 v 2
=
R2
R
Gρ 4π R 3 v 2
→
=
3R 2
R
solving for R,
⎛ 3v 2 ⎞
R=⎜
⎝ Gρ 4π ⎟⎠
1/2
2
⎡
⎤
3 ( 8.50 m/s )
=⎢
⎥
−11
2
2
3
⎢⎣ ( 6.67 × 10 N ⋅ m /kg ) ( 1 100 kg/m ) 4π ⎥⎦
1/2
= 1.53 × 10 4 m
(b)
(c)
3
4
4
ρ π R 3 = ( 1 100 kg/m 3 ) π ( 1.53 × 10 4 m ) = 1.66 × 1016 kg
3
3
4
2π R
2π R 2π ( 1.53 × 10 m )
v=
T=
=
= 1.13 × 10 4 s = 3.15 h
T
v
8.5 m/s
(d) For an illustrative model, we take your mass as 90.0 kg and
assume the asteroid is originally at rest. Angular momentum is
conserved for the asteroid-you system:
∑ Li = ∑ L f
0 = m2 vR − Iω
2
2π
0 = m2 vR − m1R 2
5
Tasteroid
4π m1R
m2 v =
5 Tasteroid
Tasteroid
16
4
4π m1R 4π ( 1.66 × 10 kg ) ( 1.53 × 10 m )
=
=
5m2 v
5 ( 90.0 kg ) ( 8.50 m/s )
= 8.37 × 1017 s = 26.5 billion years
Thus your running does not produce significant rotation of the
asteroid if it is originally stationary and does not significantly
affect any rotation it does have.
This problem is realistic. Many asteroids, such as Ida and Eros,
are roughly 30 km in diameter. They are typically irregular in
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
721
shape and not spherical. Satellites such as Phobos (of Mars),
Adrastea (of Jupiter), Calypso (of Saturn), and Ophelia (of Uranus)
would allow a visitor the same experience of easy orbital motion.
So would many Kuiper Belt objects.
P13.60
(a)
The two appropriate isolated system models are conservation
of momentum and conservation of energy applied to the system
consisting of the two spheres.
(b)
Applying conservation of momentum to the system, we find




m1v 1i + m2 v 2i = m1v 1 f + m2 v 2 f


0 + 0 = M v 1 f + 2M v 2 f


v 1 f = −2 v 2 f
(c)
Applying conservation of energy to the system, we find
K i + U i + ΔE = K f + U f
0−
−
Gm1m2
1
1
Gm1m2
+ 0 = m1 v12 f + m2 v22 f −
ri
2
2
rf
GM ( 2M ) 1
GM ( 2M )
1
= Mv12 f + ( 2M ) v22 f −
12R
2
2
4R
1
GM GM
Mv12 f =
−
− v22 f
2
2R
6R
2GM
− 2v22 f
3R
v1 f =
(d) Combining the results for parts (b) and (c),
2GM
− 2v22 f
3R
2GM
=
3R
2v2 f =
6v22 f
v2 =
P13.61
(a)
1
M
G
3
R
v1 =
2
M
G
3
R
At infinite separation U = 0 and at rest K = 0. Since the system is
isolated, the energy and momentum of the two-planet system is
conserved. We have
0=
1
1
Gm1m2
m1 v12 + m2 v22 −
2
2
d
[1]
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
722
Universal Gravitation
and
0 = m1 v1 − m2 v2
[2]
because the initial momentum of the system is zero.
Combine equations [1] and [2]:
v1 = m2
2G
d ( m1 + m2 )
and
v2 = m1
2G
d ( m1 + m2 )
The relative velocity is then
vr = v1 − ( −v2 ) =
2G ( m1 + m2 )
d
(b) The instant before the collision, the distance between the planets
is d = r1 + r2. Substitute given numerical values into the equation
found for v1 and v2 in part (a) to find
v1 = 1.03 × 10 4 m/s
and
v2 = 2.58 × 103 m/s
Therefore,
K1 =
P13.62
(a)
1
m1 v12 = 1.07 × 1032 J
2
and K 2 =
1
m2 v22 = 2.67 × 1031 J
2
The free-fall acceleration produced by the Earth is
g=
GME
= GME r −2 (directed downward)
2
r
Its rate of change is
dg
= GME ( −2 ) r −3 = −2GME r −3
dr
The minus sign indicates that g decreases with increasing height.
At the Earth’s surface,
dg
2GME
=−
dr
RE3
(b)
For small differences,
Δg
Δr
=
Δg
h
=
2GME
RE3
Thus,
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
2GME h
RE3
Δg =
(c)
Δg =
723
2 ( 6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 6.00 m )
(6.37 × 10
6
m)
3
= 1.85 × 10−5 m/s 2
P13.63
(a)
Each bit of mass dm in the ring is at the same distance from the
Gmdm
object at A. The separate contributions −
to the system
r
GmMring
energy add up to −
. When the object is at A, this is
r
−
(b)
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1 000 kg)(2.36 × 1020 kg)
(1.00 × 108 m ) + ( 2.00 × 108 m )
2
2
= −7.04 N
When the object is at the center of the ring, the potential energy of
the system is
(6.67 × 10
−
−11
N ⋅ m 2 / kg 2 ) ( 1 000 kg ) ( 2.36 × 1020 kg )
1.00 × 108 m
= −1.57 × 105 J
(c)
Total energy of the object-ring system is conserved:
(K + U ) = (K + U )
g
A
0 − 7.04 × 10 4 J =
g
B
1
1 000 kg ) vB2 − 1.57 × 105 J
(
2
⎛ 2 × 8.70 × 10 4 J ⎞
vB = ⎜
1 000 kg ⎟⎠
⎝
*P13.64
1/2
= 13.2 m/s
The original orbit radius is
r = a = 6.37 × 106 m + 500 × 103 m = 6.87 × 106 m
The original energy is
GMm
2a
(6.67 × 10−11 N ⋅ m2 kg 2 )( 5.98 × 1024 kg )(104 kg )
=−
2 ( 6.87 × 106 m )
Ei = −
= −2.90 × 1011 J
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
724
Universal Gravitation
We assume that the perigee distance in the new orbit is 6.87 × 106 m.
Then the major axis is 2a = 6.87 × 106 m + 2.00 × 107 m = 2.69 × 107 m
and the final energy is
GMm
2a
6.67 × 10−11 N ⋅ m 2 kg 2 ) ( 5.98 × 1024 kg ) ( 10 4 kg )
(
=−
2.69 × 107 m
= −1.48 × 1011 J
Ef = −
The energy input required from the engine is
E f − Ei = −1.48 × 1011 J − ( −2.90 × 1011 J ) = 1.42 × 1011 J
P13.65
From the walk, 2π r = 25 000 m. Thus, the radius of the planet is
r=
25 000 m
= 3.98 × 103 m
2π
From the drop:
Δy =
1 2 1
2
gt = g ( 29.2 s ) = 1.40 m
2
2
so,
g=
2 ( 1.40 m )
MG
−3
m/s 2 = 2
2 = 3.28 × 10
r
( 29.2 s )
which gives
M = 7.79 × 1014 kg
P13.66
The distance between the orbiting stars is
d = 2r cos 30° = 3r since cos 30° =
3
. The net
2
inward force on one orbiting star is
Gmm
GMm
cos 30° + 2
2
d
r
Gmm
mv 2
+ 2 cos 30° =
d
r
2 2
Gm2 cos 30° GM 4π r
+ 2 =
3r 2
r
rT 2
2 3
⎛ m
⎞ 4π r
G⎜
+ M⎟ =
⎝ 3
⎠
T2
ANS. FIG. P13.66
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
725
solving for the period gives
T2 =
4π 2 r 3
G M + m/ 3
(
)
⎛
⎞
r3
T = 2π ⎜
⎟
⎜⎝ G M + m/ 3 ⎟⎠
(
P13.67
(a)
1/2
)
We find the period from
15
2π r 2π ( 30 000 × 9.46 × 10 m )
T=
=
= 7 × 1015 s
5
v
2.50 × 10 m/s
= 2 × 108 yr
(b)
We estimate the mass of the Milky Way from
4π 2 ( 30 000 × 9.46 × 1015 m )
4π 2 a 3
M=
=
2 ,
GT 2
(6.67 × 10−11 N ⋅ m2 /kg 2 )(7.13 × 1015 s )
3
= 2.66 × 10 41 kg
or about 10 41 kg
Note that this is the mass of the galaxy contained within the Sun’s
orbit of the galactic center. Recent studies show that the true mass
of the galaxy, including an extended halo of dark matter, is at
least an order of magnitude larger than our estimate.
(c)
A solar mass is about 1 × 1030 kg: 1041/1030 = 1011
The number of stars is on the order of 1011 .
P13.68
Energy conservation for the two-sphere system from release to contact:
−
Gmm
Gmm 1 2 1 2
=−
+ mv + mv
R
2r
2
2
⎛ 1 1⎞
Gm ⎜ − ⎟ = v 2
⎝ 2r R ⎠
(a)
⎛
⎡ 1 1 ⎤⎞
→ v = ⎜ Gm ⎢ − ⎥⎟
⎝
⎣ 2r R ⎦⎠
1/2
The injected momentum is the final momentum of each sphere,
mv = m
2/2
⎛
⎡ 1 1 ⎤⎞
⎜⎝ Gm ⎢ 2r − R ⎥⎟⎠
⎣
⎦
1/2
⎡
⎛ 1 1⎞⎤
= ⎢Gm3 ⎜ − ⎟ ⎥
⎝ 2r R ⎠ ⎦
⎣
1/2
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
726
Universal Gravitation
(b)
If they now collide elastically each sphere reverses its velocity
to receive impulse
⎡
⎛ 1 1⎞⎤
mv − ( −mv ) = 2mv = 2 ⎢Gm3 ⎜ − ⎟ ⎥
⎝ 2r R ⎠ ⎦
⎣
P13.69
(a)
1/2
The net torque exerted on the Earth is zero. Therefore, the angular
momentum of the Earth is conserved. We use this to find the
speed at aphelion:
mra va = mrp v p
and
⎛ rp ⎞
⎛ 1.471 ⎞
4
va = v p ⎜ ⎟ = ( 3.027 × 10 4 m/s ) ⎜
⎟⎠ = 2.93 × 10 m/s
⎝
r
1.521
⎝ a⎠
(b)
Kp =
2
1
1
mv p2 = ( 5.98 × 1024 kg ) ( 3.027 × 10 4 m/s ) = 2.74 × 1033 J
2
2
Up = −
GmM
rp
(6.67 × 10
=−
−11
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 1.99 × 1030 kg )
1.471 × 1011 m
= −5.40 × 1033 J
(c)
Using the same form as in part (b),
K a = 2.57 × 1033 J and U a = −5.22 × 1033 J
(d) Compare to find that
K p + U p = −2.66 × 1033 J and K a + U a = −2.65 × 1033 J .
They agree, with a small rounding error.
P13.70
For both circular orbits,
∑ F = ma:
GME m mv 2
=
r2
r
v=
GME
r
ANS. FIG. P13.70
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
(a)
727
The original speed is
vi =
(6.67 × 10
−11
N ⋅ m 2 / kg 2 ) ( 5.98 × 1024 kg )
6.37 × 106 m + 2.00 × 105 m
= 7.79 × 103 m/s
(b)
The final speed is
vi =
(6.67 × 10
−11
N ⋅ m 2 / kg 2 ) ( 5.98 × 1024 kg )
6.47 × 106 m
= 7.85 × 103 m/s
The energy of the satellite-Earth system is
K + Ug =
(c)
1
GME m 1 GME GME
GME m
mv 2 −
= m
−
=−
2
r
2
r
r
2r
Originally,
Ei
(6.67 × 10
=−
−11
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 100 kg )
2 ( 6.57 × 106 m )
= −3.04 × 109 J
(d) Finally,
Ef
(6.67 × 10
=−
−11
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 100 kg )
2 ( 6.47 × 106 m )
= −3.08 × 109 J
(e)
Thus the object speeds up as it spirals down to the planet. The
loss of gravitational energy is so large that the total energy
decreases by
Ei − E f = −3.04 × 109 J − ( −3.08 × 109 J ) = 4.69 × 107 J
(f)
The only forces on the object are the backward force of air
resistance R, comparatively very small in magnitude, and the
force of gravity. Because the spiral path of the satellite is not
perpendicular to the gravitational force,
one component of the gravitational force pulls forward
on the satellite
to do positive work and make its speed increase.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
728
P13.71
Universal Gravitation
The centripetal acceleration of the blob comes from gravitational
acceleration:
v 2 McG 4π 2 r 2
= 2 = 2
r
r
T r
2
2 3
GMcT = 4π r
Solving for the radius gives
⎡ ( 6.67 × 10−11 N ⋅ m 2 /kg 2 )( 20 )( 1.99 × 1030 kg ) ( 5.00 × 10−3 s )2 ⎤
⎥
r=⎢
4π 2
⎢⎣
⎥⎦
1/3
rorbit = 119 km
P13.72
From Kepler’s third law, minimum period means minimum orbit size.
The “treetop satellite” in Problem 38 has minimum period. The radius
of the satellite’s circular orbit is essentially equal to the radius R of the
planet.
GMm mv 2 m ⎛ 2π R ⎞
=
= ⎜
∑ F = ma:
⎟
R2
R
R⎝ T ⎠
GρV =
2
R 2 ( 4π 2 R 2 )
RT 2
2 3
⎛4
⎞ 4π R
Gρ ⎜ π R 3 ⎟ =
⎝3
⎠
T2
The radius divides out: T 2Gρ = 3π
P13.73
→
T=
3π
Gρ
Let m represent the mass of the meteoroid and vi
its speed when far away. No torque acts on the
meteoroid, so its angular momentum is conserved
as it moves between the distant point and the
point where it grazes the Earth, moving
perpendicular to the radius:
 
 
Li = L f : mri × v i = mrf × v f
ANS. FIG. P13.73
m ( 3RE vi ) = mRE v f
v f = 3vi
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
729
Now, the energy of the meteoroid-Earth system is also conserved:
(K + U ) = (K + U )
1 2
1
GME m
mvi + 0 = mv 2f −
2
2
RE
1 2 1
GME
vi = ( 9vi2 ) −
2
2
RE
g
g
i
GME
= 4vi2 :
RE
P13.74
f
:
vi =
GME
4RE
If we choose the coordinate of the center of mass at the origin, then
0=
( Mr2 − mr1 )
M+m
and
Mr2 = mr1
(Note: this is equivalent to saying that the net torque must be zero and
the two experience no angular acceleration.) For each mass F = ma so
mr1ω 12 =
MGm
d2
and
Mr2ω 22 =
MGm
d2
ANS. FIG. P13.74
Combining these two equations and using d = r1 + r2 gives
M+m G
( r1 + r2 )ω 2 = ( 2 ) with
d
ω1 = ω2 = ω
and
T=
2π
ω
we find
T2 =
4π 2 d 3
G ( M + m)
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
730
P13.75
Universal Gravitation
The gravitational forces the particles exert on each other are in the x
direction. They do not affect the velocity of the center of mass. Energy
is conserved for the pair of particles in a reference frame coasting along
with their center of mass, and momentum conservation means that the
identical particles move toward each other with equal speeds in this
frame:
Ugi + Ki + Ki = Ugf + Kf + Kf
−
Gm1m2
Gm1m2 1
1
+0=−
+ m1v 2 + m2 v 2
ri
rf
2
2
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(1 000 kg)2
20.0 m
−11
(6.67 × 10 N ⋅ m 2 /kg 2 )(1 000 kg)2
⎛ 1⎞
=−
+ 2 ⎜ ⎟ (1 000 kg)v 2
⎝ 2⎠
2.00 m
−
⎛ 3.00 × 10−5 J ⎞
⎜⎝ 1 000 kg ⎟⎠
1/2
= v = 1.73 × 10−4 m/s
Then their vector velocities are (800 + 1.73 × 10−4 ) î m/s and
(800 − 1.73 × 10−4 )î m/s for the trailing particle and the leading
particle, respectively.
P13.76
(a)
The gravitational force exerted on m by the Earth (mass ME)
GM
accelerates m according to g 2 = 2 E . The equal-magnitude force
r
exerted on the Earth by m produces acceleration of the Earth
Gm
given by g1 = 2 . The acceleration of relative approach is then
r
g 2 + g1 =
=
Gm GME
+ 2
r2
r
(6.67 × 10−11 N ⋅ m2 /kg 2 )( 5.98 × 1024 kg + m)
(1.20 × 10
7
m)
2
⎛
⎞
m
= ( 2.77 m/s 2 ) ⎜ 1 + 24
5.98 ×10 kg ⎟⎠
⎝
(b) and (c) Here m = 5 kg and m = 2000 kg are both negligible
compared to the mass of the Earth, so the acceleration of relative
approach is just 2.77 m/s 2 .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
731
(d) Substituting m = 2.00 × 1024 kg into the expression for (g1 + g2)
above gives
g1 + g 2 = 3.70 m/s 2
(e)
Any object with mass small compared to the Earth starts to
fall with acceleration 2.77 m/s 2 . As m increases to become
comparable to the mass of the Earth, the acceleration increases,
and can become arbitrarily large. It approaches a direct proportionality to m.
P13.77
For the Earth,
∑ F = ma:
GMs m mv 2 m ⎛ 2π r ⎞
=
= ⎜
⎟
r2
r
r⎝ T ⎠
2
GMsT 2 = 4π 2 r 3
Then
Also, the angular momentum L = mvr = m
Earth. We eliminate r =
2π r
r is a constant for the
T
LT
between the equations:
2π m
⎛ LT ⎞
GMsT 2 = 4π 2 ⎜
⎝ 2π m ⎟⎠
3/2
gives
⎛ L ⎞
GMsT 1/2 = 4π 2 ⎜
⎝ 2π m ⎟⎠
3/2
Now the rates of change with time t are described by
dT ⎞
⎛1
⎛ dMs 1/2 ⎞
GMs ⎜ T −1/2
+ G⎜ 1
T ⎟ =0
⎟
⎝2
⎝ dt
⎠
dt ⎠
or
dT
dMs ⎛ T ⎞ ΔT
=−
2
≈
dt
dt ⎜⎝ Ms ⎟⎠ Δt
which gives
ΔT ≈ −Δt
dMs ⎛ T ⎞
2
dt ⎜⎝ Ms ⎟⎠
⎛ 3.16 × 107 s ⎞
9
= − ( 5 000 yr ) ⎜
⎟⎠ ( −3.64 × 10 kg/s )
1 yr
⎝
⎛
⎞
1 yr
×⎜2
30
⎝ 1.99 × 10 kg ⎟⎠
ΔT = 5.78 × 10−10 s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
732
Universal Gravitation
Challenge Problems
P13.78
Let m represent the mass of the spacecraft, rE the radius of the Earth’s
orbit, and x the distance from Earth to the spacecraft.
The Sun exerts on the spacecraft a radial inward force of
Fs =
GMs m
( rE − x )2
while the Earth exerts on it a radial outward force of
FE =
GME m
x2
The net force on the spacecraft must produce the correct centripetal
acceleration for it to have an orbital period of 1.000 year.
Thus, FS − FE =
GMSm
− x)
E
(r
which reduces to
2
−
GME m
x2
GMS
( rE − x )2
mv 2
m ⎡ 2π ( rE − x ) ⎤
=
=
⎢
⎥
( rE − x ) ( rE − x ) ⎢⎣ T ⎥⎦
2
2
GME 4π ( rE − x )
− 2 =
x
T2
[1]
Cleared of fractions, this equation would contain powers of x ranging
from the fifth to the zeroth. We do not solve it algebraically. We may
test the assertion that x is 1.48 × 109 m by substituting it into the
equation, along with the following data: MS = 1.99 × 1030 kg,
ME = 5.974 × 1024 kg, rE = 1.496 × 1011 m, and T = 1.000 yr = 3.156 × 107 s.
With x = 1.48 × 109 m, the result is
6.053 × 10−3 m/s 2 − 1.82 × 10−3 m/s 2 ≈ 5.870 8 × 10−3 m/s 2
or
5.870 9 × 10−3 m/s 2 ≈ 5.870 8 × 10−3 m/s 2
To three-digit precision, the solution is 1.48 × 109 m.
As an equation of fifth degree, equation [1] has five roots. The SunEarth system has five Lagrange points, all revolving around the Sun
synchronously with the Earth. The SOHO and ACE satellites are at
one. Another is beyond the far side of the Sun. Another is beyond the
night side of the Earth. Two more are on the Earth’s orbit, ahead of the
planet and behind it by 60°. The twin satellites of NASA’s STEREO
mission, giving three-dimensional views of the Sun from orbital
positions ahead of and trailing Earth, passed through these Lagrange
points in 2009. The Greek and Trojan asteroids are at the co-orbital
Lagrange points of the Jupiter-Sun system.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.79
(a)
From the data about perigee, the energy of the satellite-Earth
system is
E=
2
1 2 GME m 1
mv p −
= ( 1.60 kg ) ( 8.23 × 103 m/s )
2
rp
2
−
or
(b)
733
(6.67 × 10
−11
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 1.60 kg )
7.02 × 106 m
E = −3.67 × 107 J
L = mvr sin θ = mv p rp sin 90.0°
= ( 1.60 kg ) ( 8.23 × 103 m/s ) ( 7.02 × 106 m )
= 9.24 × 1010 kg ⋅ m 2 /s
(c)
Since both the energy of the satellite-Earth system and the
angular momentum of the Earth are conserved,
at apogee we must have
1 2 GMm
mva −
=E
2
ra
and mva ra sin 90.0° = L
Thus,
1
(1.60 kg ) va2
2
6.67 × 10−11 N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg ) ( 1.60 kg )
(
−
ra
= −3.67 × 107 J
and
(1.60 kg ) va ra = 9.24 × 1010 kg ⋅ m2 /s
Solving simultaneously, and suppressing units,
(6.67 × 10
1
(1.60) va2 −
2
−11
)( 5.98 × 10 )(1.60)(1.60) v
24
9.24 × 10
a
10
= −3.67 × 107
which reduces to
0.800va2 − 11 046va + 3.672 3 × 107 = 0
so
va =
11 046 ±
(11 046)2 − 4 ( 0.800)( 3.672 3 × 107 )
2 ( 0.800 )
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
734
Universal Gravitation
This gives va = 8 230 m/s or 5 580 m/s . The smaller answer
refers to the velocity at the apogee while the larger refers to
perigee.
Thus,
9.24 × 1010 kg ⋅ m 2 /s
L
=
= 1.04 × 107 m
mva ( 1.60 kg ) ( 5.58 × 103 m/s )
ra =
(d) The major axis is 2a = rp + ra, so the semimajor axis is
a=
(e)
T=
1
(7.02 × 106 m + 1.04 × 107 m ) = 8.69 × 106 m
2
4π 2 a 3
=
GME
(6.67 × 10
4π 2 ( 8.69 × 106 m )
−11
3
N ⋅ m 2 /kg 2 ) ( 5.98 × 1024 kg )
T = 8 060 s = 134 min
*P13.80
(a)
Energy of the spacecraft-Mars system is conserved as the
spacecraft moves between a very distant point and the point of
closest approach:
1
GMMars m
mvr2 −
2
r
2GMMars
r
0+0=
vr =
After the engine burn, for a circular orbit we have
GMMars m mv02
=
r2
r
∑ F = ma:
v0 =
GMMars
r
The percentage reduction from the original speed is
vr − v0
=
vr
(b)
2v0 − v0
=
2v0
2 −1
× 100% = 29.3%
2
The answer to part (a) applies with no changes , as the solution
to part (a) shows.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
735
ANSWERS TO EVEN-NUMBERED PROBLEMS
P13.2
~10−7 N
P13.4
(a) 4.39 × 1020 N; (b) 1.99 × 1020 N; (c) 3.55 × 1022 N; (d) The force exerted
by the Sun on the Moon is much stronger than the force of the Earth on
the Moon.
P13.6
( −10.0î + 5.93 ĵ) × 10
−11
N
P13.8
The situation is impossible because no known element could compose
the spheres.
P13.10
3.06 × 10–8 m
P13.12
2
3
P13.14
(a)
(r
2MGr
2
+ a2 )
3/2
toward the center of mass; (b) At r = 0, the fields of the
two objects are equal in magnitude and opposite in direction, to add to
zero; (c) As r → 0, 2MGr ( r 2 + a 2 )
approaches 2MG(0)/a3 = 0; (d)
When r is much greater than a, the angles the field vectors make with
the x axis become smaller. At very great distances, the field vectors are
almost parallel to the axis; therefore they begin to look like the field
vector from a single object of mass 2M; (e) As r becomes much larger
than a, the expression approaches
−3/2
2MGr ( r 2 + 02 )
−3/2
= 2MGr/r 3 = 2MG/r 2 as required.
P13.16
(a) 1.31 × 1017 N; (b) 2.62 × 1012 N/kg
P13.18
1.90 × 1027 kg
P13.20
(a) The particle does posses angular momentum because it is not
headed straight for the origin. (b) Its angular momentum is constant.
There are no identified outside influences acting on the object. (c) See
P13.20(c) for full explanation.
P13.22
1.30 revolutions
P13.24
1.27
P13.26
1.63 × 104 rad/s
P13.28
(a) 6.02 × 10 kg; (b) The Earth wobbles a bit as the Moon orbits it, so
both objects move nearly in circles about their center of mass, staying
on opposite sides of it.
24
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
736
Universal Gravitation
P13.30
(a) −4.77 × 109 J; (b) 569 N
P13.32
4.17 × 1010 J
P13.34
(a) See P13.34 for full description; (b) 340 s
P13.36
1.66 × 104 m/s
P13.38
2v
P13.40
(a) 0.980; (b) 127 yr; (c) –2.13 × 1017 J
P13.42
GME m
12RE
P13.44
(a) 42.1 km/s; (b) 2.20 × 1011 m
( RE + h)3 ; (b)
⎡ RE + 2h ⎤
2π 2 RE2 m
GME
; (c) GME m ⎢
−
⎥
2
RE + h
⎣ 2RE ( RE + h ) ⎦ ( 86 400 s )
P13.46
(a) 2π
P13.48
⎛ GME ⎞
(a) vi = ⎜
⎝ r ⎟⎠
P13.50
(a) 3.07 × 106 m; (b) the rocket would travel farther from Earth
P13.52
492 m/s
P13.54
If one uses the result v =
P13.56
(a) 0.700 rad/s; (b) Because his feet stay in place on the floor, his head
will be moving at the same tangential speed as his feet. However, his
feet and his head are travelling in circles of different radii; (c) If he’s
not careful, there could be a collision between his head and the wall
(see P13.56 for full explanation)
P13.58
(a) h =
GME
1/2
5 GME ⎞
; (b) ⎛⎜
⎟
4⎝ r ⎠
1/2
; (c) rf =
25r
7
GM
and the relation v = (2πτ /T), one finds
r
the radius of the orbit to be smaller than the radius of the Earth, so the
spacecraft would need to be in orbit underground.
RE vi2
h
; (b) v f = vesc
; (c) With
2
2
vesc − vi
RE + h
v1 << vesc , h ≈
GME
RE vi2 RE vi2 RE
vi2
g
=
,
But
so
in agreement
=
.
h
=
2
RE2
vesc
2GME
2g
with 02 = vi2 + 2 ( − g ) ( h − 0 ) .
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 13
P13.60
(a) The two appropriate isolated system models are conservation of
momentum and conservation of energy applied to the system
2GM

− 2v22 f ;
consisting of the two spheres; (b) −2 v 2 f ; (c)
3R
1
M
2
M
G
G
(d) v2 =
, v1 =
3
R
3
R
P13.62
(a)
P13.64
1.42 × 1011 J
P13.66
See P13.66 for the full answer.
P13.68
⎡
⎛ 1 1⎞⎤
(a) mv = ⎢GM 3 ⎜ − ⎟ ⎥
⎝ 2r R ⎠ ⎦
⎣
737
dg
2GME
2GME h
=−
; (b) Δg =
; (c) 1.85 × 10−5 m/s2
3
3
dr
RE
RE
1/2
⎡
⎛ 1 1⎞⎤
; (b) 2 ⎢GM 3 ⎜ − ⎟ ⎥
⎝ 2r R ⎠ ⎦
⎣
1/2
P13.70
(a) 7.79 × 103 m/s; (b) 7.85 × 103 m/s; (c) −3.04 × 109 J; (d) −3.08 × 109 J;
(e) 4.69 × 107 J; (f) one component of the gravitational force pulls
forward on the satellite
P13.72
See P13.72 for full description.
P13.74
See P13.74 for full description.
P13.76
⎛
⎞
m
(a) ( 2.77 m/s 2 ) ⎜ 1 +
; (b and c) 2.77 m/s2; (d) 3.70 m/s2;
24
⎟
5.98 × 10 kg ⎠
⎝
(e) Any object with mass small compared to the Earth starts to fall with
acceleration 2.77 m/s2. As m increases to become comparable to the
mass of the Earth, the acceleration increases and can become arbitrarily
large. It approaches a direct proportionality to m.
P13.78
See P13.78 for full description.
P13.80
(a) 29.3%; (b) no changes
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.